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Question Number 52703 by maxmathsup by imad last updated on 11/Jan/19

l e t f (t) = \int_{0}^{\infty} \frac{{c o s}^{2} (t x)}{{(x^{2} + 3)}^{2}} d x w i t h t ⩾ 0

1) g i v e a e x p l i c i t f o r m o f f (t)

2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{x s i n (2 t x)}{{(x^{2} + 3)}^{2}} d x

3) g i v e t h e v a l u e s o f i n t e g r a l s \int_{0}^{\infty} \frac{d x}{{(x^{2} + 3)}^{2}} a n d \int_{0}^{\infty} \frac{{c o s}^{2} (π x)}{{(x^{2} + 3)}^{2}} d x

4) g i v e t h e v a l u e s o f i n t e g r a l s \int_{0}^{\infty} \frac{x s i n (π x)}{{(x^{2} + 3)}^{2}} a n d \int_{0}^{\infty} \frac{x s i n (\frac{π x}{2})}{{(x^{2} + 3)}^{2}} d x .

Commented by Abdo msup. last updated on 13/Jan/19

c h a n n g e m e n t x = \sqrt{3} u g i v e

f (t) = \int_{0}^{\infty} \frac{{c o s}^{2} (\sqrt{3} t u)}{9 {(u^{2} + 1)}^{2}} \sqrt{3} d u

= \frac{\sqrt{3}}{9} \int_{0}^{+ \infty} \frac{{c o s}^{2} (\sqrt{3} t u)}{{(u^{2} + 1)}^{2}} d u = \frac{\sqrt{3}}{18} \int_{- \infty}^{+ \infty} \frac{{c o s}^{2} (\sqrt{3} t u)}{{(u^{2} + 1)}^{2}} d u

= \frac{\sqrt{3}}{18} \int_{- \infty}^{+ \infty} \frac{1 + c o s (2 \sqrt{3} t u)}{2 {(u^{2} + 1)}^{2}} d u

= \frac{\sqrt{3}}{36} \int_{- \infty}^{+ \infty} \frac{d u}{{(u^{2} + 1)}^{2}} + \frac{\sqrt{3}}{36} \int_{- \infty}^{+ \infty} \frac{c o s (2 \sqrt{3} t u) d u}{{(u^{2} + 1)}^{2}}

\int_{- \infty}^{+ \infty} \frac{d u}{{(u^{2} + 1)}^{2}} =_{u = t a n θ} \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 + {t a n}^{2} θ}{{(1 + {t a n}^{2} θ)}^{2}} d θ

= 2 \int_{0}^{\frac{π}{2}} {c o s}^{2} θ d θ = 2 \int_{0}^{\frac{π}{2}} \frac{1 + c o s (2 θ)}{2} d θ

= \frac{π}{2} + {[\frac{1}{2} s i n (2 θ)]}_{0}^{\frac{π}{2}} = \frac{π}{2} l e t d e t e r m i n e

\int_{- \infty}^{+ \infty} \frac{c o s (2 \sqrt{3} t u)}{{(u^{2} + 1)}^{2}} d u = I

I = R e (\int_{- \infty}^{+ \infty} \frac{e^{2 i \sqrt{3} t u}}{{(u^{2} + 1)}^{2}} d u) l e t

φ (z) = \frac{e^{2 i \sqrt{3} t z}}{{(z^{2} + 1)}^{2}} \Rightarrow φ (z) = \frac{e^{2 i \sqrt{3} t z}}{{(z - i)}^{2} {(z + i)}^{2}}

t h e p o l e s o f φ a r e i a n d - i (d o u b l e s) s o

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e (φ, i)

R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}

= {l i m}_{z \to i} {\frac{e^{2 i \sqrt{3} t z}}{{(z + i)}^{2}}}^{(1)}

= {l i m}_{z \to i} \frac{2 i \sqrt{3} t e^{2 i \sqrt{3} t z} {(z + i)}^{2} - 2 (z + i) e^{2 i \sqrt{3} t z}}{{(z + i)}^{4}}

= {l i m}_{z \to i} \frac{(2 i \sqrt{3} t (z + i) - 2) e^{2 i \sqrt{3} t z}}{{(z + i)}^{3}}

= \frac{(2 i \sqrt{3} t (2 i) - 2) e^{- 2 \sqrt{3} t}}{{(2 i)}^{3}} = \frac{(- 4 i \sqrt{3} t - 2) e^{- 2 \sqrt{3} t}}{- 8 i}

= \frac{(2 i \sqrt{3} t + 1) e^{- 2 \sqrt{3} t}}{4 i} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z =

2 i π \frac{(2 i \sqrt{3} t + 1) e^{- 2 \sqrt{3} t}}{4 i} = \frac{π}{2} (2 i \sqrt{3} t + 1) e^{- 2 \sqrt{3} t} b u t

I = R e (\int_{- \infty}^{+ \infty} φ (z) d z) = \frac{π}{2} e^{- 2 \sqrt{3} t} \Rightarrow

f (t) = \frac{π \sqrt{3}}{72} + \frac{\sqrt{3}}{36} {\frac{π}{2} e^{- 2 \sqrt{3} t}} \Rightarrow

f (t) = \frac{π \sqrt{3}}{72} (1 + e^{- 2 \sqrt{3} t}) .